You can get todays day time and date using this code
<?php
echo date("d")." ";
echo date("m")." ";
echo date("Y")." ";
echo date("h:i:s A");
ECHO ' <br/>';
echo jddayofweek ( cal_to_jd(CAL_GREGORIAN, date("m"),date("d"), date("Y")) , 1 );
?>
JDDayOfWeek
(PHP 4, PHP 5)
JDDayOfWeek — Vracia deň v týždni
Popis
mixed jddayofweek
( int $julianskyden
, int $mod
)
Vracia deň v týždni. Môže vrátiť reťazec alebo integer závisiac na móde.
| Mód | Význam |
|---|---|
| 0 | Vracia číslo dňa ako integer (0=nedeľa, 1=pondelok, atď) |
| 1 | Vracia reťazec obsahujúci deň v týždni (anglicko-gregoriánsky) |
| 2 | Vracia reťazec obsahujúci skrátený tvar dňa v týždni (anglicko-gregoriánsky) |
JDDayOfWeek
nrkkalyan at rediffmail dot com
25-Feb-2005 10:08
25-Feb-2005 10:08
php at xtramicro dot com
07-Sep-2004 08:28
07-Sep-2004 08:28
Be aware that date() and mktime() only work as long as you move within the UNIX era (1970 - 2038 / 0x0 - 0x7FFFFFFF in seconds). Outside that era those functions are only generating errors.
In other words: mktime(0, 0, 0, 12, 31, 1969) *DOES NOT* work (and so doesn't date() fed with with mktime()'s result from above). But cal_to_jd(CAL_GREGORIAN, 12, 11, 1969) *DOES WORK*.
And please note that the calendar-extension's functions arguments follow the US date order: month - day - year.
